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3 years ago

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A man who works for a moving company is loading a box onto a moving van. He pushes a 200N box up a 5m long ramp. If he pushes wi

th a force of 60N and the ramp is 1m high, what is the efficiency of the inclined plane

1 answer:

OlgaM077 [116]3 years ago

8 0

Answer:

η = 0.667 = 66.7%

Explanation:

The efficiency of the man can be given by the following formula:

η = output/input

where,

η = efficiency of man = ?

output = potential energy gain of the box = Wh

input = work done by man = Fd

Therefore,

$\eta = \frac{Wh}{Fd}$

where,

W = weight of box = 200 N

h = height gained by box = 1 m

F = force exerted by man = 60 N

d = length of ramp = 5 m

Therefore,

$\eta = \frac{(200\ N)(1\ m)}{(60\ N)(5\ m)}$

<u>η = 0.667 = 66.7%</u>

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The magnitude of the gravitational field strength near Earth's surface is represented by

Zanzabum

Answer:

The magnitude of the gravitational field strength near Earth's surface is represented by approximately $9.82\,\frac{m}{s^{2}}$ .

Explanation:

Let be M and m the masses of the planet and a person standing on the surface of the planet, so that M >> m. The attraction force between the planet and the person is represented by the Newton's Law of Gravitation:

$F = G\cdot \frac{M\cdot m}{r^{2}}$

Where:

$M$ - Mass of the planet Earth, measured in kilograms.

$m$ - Mass of the person, measured in kilograms.

$r$ - Radius of the Earth, measured in meters.

$G$ - Gravitational constant, measured in $\frac{m^{3}}{kg\cdot s^{2}}$ .

But also, the magnitude of the gravitational field is given by the definition of weight, that is:

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Where:

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$g$ - Gravity constant, measured in meters per square second.

After comparing this expression with the first one, the following equivalence is found:

$g = \frac{G\cdot M}{r^{2}}$

Given that $G = 6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}}$ , $M = 5.972 \times 10^{24}\,kg$ and $r = 6.371 \times 10^{6}\,m$ , the magnitude of the gravitational field near Earth's surface is:

$g = \frac{\left(6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}} \right)\cdot (5.972\times 10^{24}\,kg)}{(6.371\times 10^{6}\,m)^{2}}$

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Suppose that you'd like to find out if a distant star is moving relative to the earth. The star is much too far away to detect a

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Answer:

The speed will be "18km/s". A further explanation is given below.

Explanation:

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Wavelength,

$\lambda = 656.46 \ nm$

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As we know,

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On substituting the values, we get

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⇒ $=18 \ km/s$

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